All Answers 358059 Answers

The probability is 6/25.

The area of the square is 15.

The answer is (-2,7).

1111 

211

121

112

22

I can't think of any more.

(a) The left-hand side counts the number of ways to choose k elements from a set of n - 1 elements. The first term, C(n - 1, k - 1), counts the number of ways to choose k - 1 elements, and the second term, C(n - 1, k), counts the number of ways to choose k elements. To form the set of k elements in C(n, k), we simply add an element n to each of these two sets. So, C(n - 1, k - 1) + C(n - 1, k) = C(n, k).

(b) The left-hand side counts the number of ways to choose 0 to n elements from a set of n elements. For each positive integer i from 0 to n, C(n, i) counts the number of ways to choose i elements. The sum on the left-hand side is equivalent to counting the number of possible combinations of n elements, which is 2^n.

(c) The left-hand side counts the number of ways to choose 0 to k elements from a set of n + k elements. Each term C(n + i, i) counts the number of ways to choose i elements from a set of n + i elements. The sum on the left-hand side is equivalent to counting the number of possible combinations of n + k elements, which is C(n + k + 1, k).

(d) The left-hand side counts the number of ordered pairs (X, Y) where X is a set of r elements selected from a set of m elements and Y is a set of r elements selected from a set of n elements. For each positive integer i from 0 to r, C(m, i) C(n, r - i) counts the number of ordered pairs where X has i elements and Y has r - i elements. The sum on the left-hand side is equivalent to counting the number of possible combinations of r elements from the set of m + n elements, which is C(m + n, r).

(e) The left-hand side counts the number of ordered pairs (a, b) of positive integers where 1 <= a, b <= n. The expression (n + 1)^2 - 2n - 1 counts the total number of ordered pairs of positive integers where 1 <= a, b <= n + 1 and then subtracts the ordered pairs (n + 1, n + 1) and (n + 1, i) for 1 <= i <= n, as well as (i, n + 1) for 1 <= i <= n. This gives us the total number of ordered pairs of positive integers where 1 <= a, b <= n.

(f) The left-hand side counts the number of ordered triples (a, b, c) of positive integers where 1 <= a, b, c <= n. The expression (n + 1)^3 - 3n^2 - 3n - 1 counts the total number of ordered triples of positive integers where 1 <= a, b, c <= n + 1 and then subtracts the ordered triples (n + 1, n + 1, n + 1), (n + 1, n + 1, i), (n + 1, i, n + 1), and (i, n + 1, n + 1) for 1 <= i <= n, as well as all ordered triples where at least one element is equal to n + 1. This gives us the total number of ordered triples of positive integers where 1 <= a, b, c <= n.

(a) The left-hand side counts the number of ways to choose k elements from a set of n - 1 elements. The first term, C(n - 1, k - 1), counts the number of ways to choose k - 1 elements, and the second term, C(n - 1, k), counts the number of ways to choose k elements. To form the set of k elements in C(n, k), we simply add an element n to each of these two sets. So, C(n - 1, k - 1) + C(n - 1, k) = C(n, k).

(b) The left-hand side counts the number of ways to choose 0 to n elements from a set of n elements. For each positive integer i from 0 to n, C(n, i) counts the number of ways to choose i elements. The sum on the left-hand side is equivalent to counting the number of possible combinations of n elements, which is 2^n.

(c) The left-hand side counts the number of ways to choose 0 to k elements from a set of n + k elements. Each term C(n + i, i) counts the number of ways to choose i elements from a set of n + i elements. The sum on the left-hand side is equivalent to counting the number of possible combinations of n + k elements, which is C(n + k + 1, k).

(d) The left-hand side counts the number of ordered pairs (X, Y) where X is a set of r elements selected from a set of m elements and Y is a set of r elements selected from a set of n elements. For each positive integer i from 0 to r, C(m, i) C(n, r - i) counts the number of ordered pairs where X has i elements and Y has r - i elements. The sum on the left-hand side is equivalent to counting the number of possible combinations of r elements from the set of m + n elements, which is C(m + n, r).

(e) The left-hand side counts the number of ordered pairs (a, b) of positive integers where 1 <= a, b <= n. The expression (n + 1)^2 - 2n - 1 counts the total number of ordered pairs of positive integers where 1 <= a, b <= n + 1 and then subtracts the ordered pairs (n + 1, n + 1) and (n + 1, i) for 1 <= i <= n, as well as (i, n + 1) for 1 <= i <= n. This gives us the total number of ordered pairs of positive integers where 1 <= a, b <= n.

(f) The left-hand side counts the number of ordered triples (a, b, c) of positive integers where 1 <= a, b, c <= n. The expression (n + 1)^3 - 3n^2 - 3n - 1 counts the total number of ordered triples of positive integers where 1 <= a, b, c <= n + 1 and then subtracts the ordered triples (n + 1, n + 1, n + 1), (n + 1, n + 1, i), (n + 1, i, n + 1), and (i, n + 1, n + 1) for 1 <= i <= n, as well as all ordered triples where at least one element is equal to n + 1. This gives us the total number of ordered triples of positive integers where 1 <= a, b, c <= n.

The answer is 20*sqrt(3).

Note that cos(22) = √(1 - k^2)  from the previous result.

Also cos(22) = cos(11 + 11) = 1 - 2*sin(11)^2

Can you take it from there?

Hi Alan,

Yes, I worked on it and kinda got it right, thank you klindly, however, maybe just one that I do not see how to go about is to calculate Sin11?...Must I devide Sin22 by 2 and then proceed, in which case I'm not sure how to?..or is it an entirely different approach?..please if you don't mind..

cos(158) = cos(180 - 22) = cos(180)*cos(22) + sin(180)*sin(22) = -cos(22) = -√(1-sin(22)^2) = - √(1 - k^2)

Let O be the origin of a 3D co-ordinate system with OA, OB and OC lying along the Ox, Oy and Oz axes respectively.

The equation of the ABC plane will be

\(\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1.\)

If h is the side length of the cube, then the co-ordinates of the point on the cube furthest from the origin will be

(h, h, h), and this point is to lie on the plane, in which case

\(\displaystyle \frac{h}{a}+\frac{h}{b}+\frac{h}{c}=1,\)

so

\(\displaystyle h\left\{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right\}=1 \\ \displaystyle h = \frac{abc}{ab+bc+ca}.\)

We have &(12&)=&(7-12)=&(-5)= (-5-7)=-12

10 - 99 : 11
100 - 999 : 112
1000 - 9999 : 1124
10000 - 99999 : 11248
100,000 - 999,999 : 112,496
1000000 - 9999999 : 1124992
10000000 - 99999999 : 11249985
100000000 - 999999999 : 112499976
1000000000 - 9999999999 : 1124999972

Upload your diagram !!!!!

C)

n=1;a=99;s=0;m=n;p=0;n=m;cycle:s=s+(n%10);p=p+1;n=(n div 10);if(n!=0, goto cycle,0);m=m+1;if(m<=a, goto5,0);print"Total Sum =",s;print"Total Num =",p

Total Sum = 900
Total Num = 189

Since G is the centroid of triangle ABC, we know that it is the intersection point of the medians of the triangle. A median divides a triangle into two smaller triangles of equal area. In this case, we have triangle ABG as one of those smaller triangles, and it's given that ABG is an equilateral triangle with side length 2.

The area of an equilateral triangle can be found using the formula:

Area = (side^2 * √3) / 4

For triangle ABG, we have:

Area(ABG) = (2^2 * √3) / 4 = 4√3 / 4 = √3

Since triangle ABC is composed of 3 smaller triangles with equal area (ABG, ACG, and BCG), we can find the area of triangle ABC by multiplying the area of triangle ABG by 3:

Area(ABC) = 3 * Area(ABG) = 3 * √3 = 3√3

So the area of triangle ABC is 3√3 square units.

3.2 means: 3 + 2/10 ==[10 * 3  +  2] ==32 / 10............(1)

0.44 means: 44 / 100...............(2)

Multiply (1) by (2): 32 / 10  *  44 / 100 ==[32  *  44] / [10 * 100] ==1,408 / 1,000 ==1.408

(a) The left-hand side counts the number of ways to choose k elements from a set of n - 1 elements. The first term, C(n - 1, k - 1), counts the number of ways to choose k - 1 elements, and the second term, C(n - 1, k), counts the number of ways to choose k elements. To form the set of k elements in C(n, k), we simply add an element n to each of these two sets. So, C(n - 1, k - 1) + C(n - 1, k) = C(n, k).

(b) The left-hand side counts the number of ways to choose 0 to n elements from a set of n elements. For each positive integer i from 0 to n, C(n, i) counts the number of ways to choose i elements. The sum on the left-hand side is equivalent to counting the number of possible combinations of n elements, which is 2^n.

(c) The left-hand side counts the number of ways to choose 0 to k elements from a set of n + k elements. Each term C(n + i, i) counts the number of ways to choose i elements from a set of n + i elements. The sum on the left-hand side is equivalent to counting the number of possible combinations of n + k elements, which is C(n + k + 1, k).

(d) The left-hand side counts the number of ordered pairs (X, Y) where X is a set of r elements selected from a set of m elements and Y is a set of r elements selected from a set of n elements. For each positive integer i from 0 to r, C(m, i) C(n, r - i) counts the number of ordered pairs where X has i elements and Y has r - i elements. The sum on the left-hand side is equivalent to counting the number of possible combinations of r elements from the set of m + n elements, which is C(m + n, r).

(e) The left-hand side counts the number of ordered pairs (a, b) of positive integers where 1 <= a, b <= n. The expression (n + 1)^2 - 2n - 1 counts the total number of ordered pairs of positive integers where 1 <= a, b <= n + 1 and then subtracts the ordered pairs (n + 1, n + 1) and (n + 1, i) for 1 <= i <= n, as well as (i, n + 1) for 1 <= i <= n. This gives us the total number of ordered pairs of positive integers where 1 <= a, b <= n.

(f) The left-hand side counts the number of ordered triples (a, b, c) of positive integers where 1 <= a, b, c <= n. The expression (n + 1)^3 - 3n^2 - 3n - 1 counts the total number of ordered triples of positive integers where 1 <= a, b, c <= n + 1 and then subtracts the ordered triples (n + 1, n + 1, n + 1), (n + 1, n + 1, i), (n + 1, i, n + 1), and (i, n + 1, n + 1) for 1 <= i <= n, as well as all ordered triples where at least one element is equal to n + 1. This gives us the total number of ordered triples of positive integers where 1 <= a, b, c <= n.

(a) The left-hand side counts the number of ways to choose k elements from a set of n - 1 elements. The first term, C(n - 1, k - 1), counts the number of ways to choose k - 1 elements, and the second term, C(n - 1, k), counts the number of ways to choose k elements. To form the set of k elements in C(n, k), we simply add an element n to each of these two sets. So, C(n - 1, k - 1) + C(n - 1, k) = C(n, k).

(b) The left-hand side counts the number of ways to choose 0 to n elements from a set of n elements. For each positive integer i from 0 to n, C(n, i) counts the number of ways to choose i elements. The sum on the left-hand side is equivalent to counting the number of possible combinations of n elements, which is 2^n.

(c) The left-hand side counts the number of ways to choose 0 to k elements from a set of n + k elements. Each term C(n + i, i) counts the number of ways to choose i elements from a set of n + i elements. The sum on the left-hand side is equivalent to counting the number of possible combinations of n + k elements, which is C(n + k + 1, k).

(d) The left-hand side counts the number of ordered pairs (X, Y) where X is a set of r elements selected from a set of m elements and Y is a set of r elements selected from a set of n elements. For each positive integer i from 0 to r, C(m, i) C(n, r - i) counts the number of ordered pairs where X has i elements and Y has r - i elements. The sum on the left-hand side is equivalent to counting the number of possible combinations of r elements from the set of m + n elements, which is C(m + n, r).

(e) The left-hand side counts the number of ordered pairs (a, b) of positive integers where 1 <= a, b <= n. The expression (n + 1)^2 - 2n - 1 counts the total number of ordered pairs of positive integers where 1 <= a, b <= n + 1 and then subtracts the ordered pairs (n + 1, n + 1) and (n + 1, i) for 1 <= i <= n, as well as (i, n + 1) for 1 <= i <= n. This gives us the total number of ordered pairs of positive integers where 1 <= a, b <= n.

(f) The left-hand side counts the number of ordered triples (a, b, c) of positive integers where 1 <= a, b, c <= n. The expression (n + 1)^3 - 3n^2 - 3n - 1 counts the total number of ordered triples of positive integers where 1 <= a, b, c <= n + 1 and then subtracts the ordered triples (n + 1, n + 1, n + 1), (n + 1, n + 1, i), (n + 1, i, n + 1), and (i, n + 1, n + 1) for 1 <= i <= n, as well as all ordered triples where at least one element is equal to n + 1. This gives us the total number of ordered triples of positive integers where 1 <= a, b, c <= n.

Think about it, is this even possible?

(a) The numbers from 1 to 99 include 9 one-digit numbers (1 to 9), 90 two-digit numbers (10 to 99), and a total of 99 numbers. The one-digit numbers have 1 digit each, the two-digit numbers have 2 digits each. Therefore, the total number of digits is:

9 × 1 + 90 × 2 = 9 + 180 = 189

Sam wrote down 189 digits in total.

(b) There are no 0's among the one-digit numbers. Among the two-digit numbers, there are 10 numbers that have 0 as the second digit (10, 20, 30, ..., 90). Therefore, out of the 99 numbers Sam wrote down, 10 have 0 as the second digit. The probability of choosing one of these numbers is:

P(choosing 0) = 10/99

(c) We can calculate the sum of the digits by adding the sum of the digits in each of the numbers. There are 9 one-digit numbers, which add up to 45. The two-digit numbers can be divided into two groups: those whose first digit is 1 (10 to 19) and those whose first digit is 2 to 9 (20 to 99). The sum of the digits in the first group is:

(1 + 0) + (1 + 1) + (1 + 2) + ... + (1 + 9) = 55

To calculate the sum of the digits in the second group, we first notice that there are 8 numbers whose first digit is 2, and each of these numbers contributes 2 to the sum of the digits. Similarly, there are 8 numbers whose first digit is 3, and each of these numbers contributes 3 to the sum of the digits, and so on up to 9 numbers whose first digit is 9, and each of these numbers contributes 9 to the sum of the digits. Therefore, the sum of the digits in the second group is:

8 × 2 + 8 × 3 + ... + 9 × 9 = 2(8 + 9) + 3(8 + 9) + ... + 9(8 + 9) = (8 + 9)(2 + 3 + ... + 9) = 45 × 44/2 = 990

Therefore, the total sum of the digits that Sam wrote down is:

45 + 55 + 990 = 1090

(a) The left-hand side counts the number of ways to choose k elements from a set of n - 1 elements. The first term, C(n - 1, k - 1), counts the number of ways to choose k - 1 elements, and the second term, C(n - 1, k), counts the number of ways to choose k elements. To form the set of k elements in C(n, k), we simply add an element n to each of these two sets. So, C(n - 1, k - 1) + C(n - 1, k) = C(n, k).

(b) The left-hand side counts the number of ways to choose 0 to n elements from a set of n elements. For each positive integer i from 0 to n, C(n, i) counts the number of ways to choose i elements. The sum on the left-hand side is equivalent to counting the number of possible combinations of n elements, which is 2^n.

(c) The left-hand side counts the number of ways to choose 0 to k elements from a set of n + k elements. Each term C(n + i, i) counts the number of ways to choose i elements from a set of n + i elements. The sum on the left-hand side is equivalent to counting the number of possible combinations of n + k elements, which is C(n + k + 1, k).

(d) The left-hand side counts the number of ordered pairs (X, Y) where X is a set of r elements selected from a set of m elements and Y is a set of r elements selected from a set of n elements. For each positive integer i from 0 to r, C(m, i) C(n, r - i) counts the number of ordered pairs where X has i elements and Y has r - i elements. The sum on the left-hand side is equivalent to counting the number of possible combinations of r elements from the set of m + n elements, which is C(m + n, r).

(e) The left-hand side counts the number of ordered pairs (a, b) of positive integers where 1 <= a, b <= n. The expression (n + 1)^2 - 2n - 1 counts the total number of ordered pairs of positive integers where 1 <= a, b <= n + 1 and then subtracts the ordered pairs (n + 1, n + 1) and (n + 1, i) for 1 <= i <= n, as well as (i, n + 1) for 1 <= i <= n. This gives us the total number of ordered pairs of positive integers where 1 <= a, b <= n.

(f) The left-hand side counts the number of ordered triples (a, b, c) of positive integers where 1 <= a, b, c <= n. The expression (n + 1)^3 - 3n^2 - 3n - 1 counts the total number of ordered triples of positive integers where 1 <= a, b, c <= n + 1 and then subtracts the ordered triples (n + 1, n + 1, n + 1), (n + 1, n + 1, i), (n + 1, i, n + 1), and (i, n + 1, n + 1) for 1 <= i <= n, as well as all ordered triples where at least one element is equal to n + 1. This gives us the total number of ordered triples of positive integers where 1 <= a, b, c <= n.

(a) The left-hand side counts the number of ways to choose k elements from a set of n - 1 elements. The first term, C(n - 1, k - 1), counts the number of ways to choose k - 1 elements, and the second term, C(n - 1, k), counts the number of ways to choose k elements. To form the set of k elements in C(n, k), we simply add an element n to each of these two sets. So, C(n - 1, k - 1) + C(n - 1, k) = C(n, k).

(b) The left-hand side counts the number of ways to choose 0 to n elements from a set of n elements. For each positive integer i from 0 to n, C(n, i) counts the number of ways to choose i elements. The sum on the left-hand side is equivalent to counting the number of possible combinations of n elements, which is 2^n.

(c) The left-hand side counts the number of ways to choose 0 to k elements from a set of n + k elements. Each term C(n + i, i) counts the number of ways to choose i elements from a set of n + i elements. The sum on the left-hand side is equivalent to counting the number of possible combinations of n + k elements, which is C(n + k + 1, k).

(d) The left-hand side counts the number of ordered pairs (X, Y) where X is a set of r elements selected from a set of m elements and Y is a set of r elements selected from a set of n elements. For each positive integer i from 0 to r, C(m, i) C(n, r - i) counts the number of ordered pairs where X has i elements and Y has r - i elements. The sum on the left-hand side is equivalent to counting the number of possible combinations of r elements from the set of m + n elements, which is C(m + n, r).

(e) The left-hand side counts the number of ordered pairs (a, b) of positive integers where 1 <= a, b <= n. The expression (n + 1)^2 - 2n - 1 counts the total number of ordered pairs of positive integers where 1 <= a, b <= n + 1 and then subtracts the ordered pairs (n + 1, n + 1) and (n + 1, i) for 1 <= i <= n, as well as (i, n + 1) for 1 <= i <= n. This gives us the total number of ordered pairs of positive integers where 1 <= a, b <= n.

(f) The left-hand side counts the number of ordered triples (a, b, c) of positive integers where 1 <= a, b, c <= n. The expression (n + 1)^3 - 3n^2 - 3n - 1 counts the total number of ordered triples of positive integers where 1 <= a, b, c <= n + 1 and then subtracts the ordered triples (n + 1, n + 1, n + 1), (n + 1, n + 1, i), (n + 1, i, n + 1), and (i, n + 1, n + 1) for 1 <= i <= n, as well as all ordered triples where at least one element is equal to n + 1. This gives us the total number of ordered triples of positive integers where 1 <= a, b, c <= n.